A=\(\dfrac{2^{13}\times9^5}{6^8\times8^3}\)
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\(C=\dfrac{5\times2^{12}\times3^8-3^9\times2^{12}}{2^2\times2^{13}\times3^8+2\times2^{12}\times\left(-3^9\right)}=\dfrac{3^8\times2^{12}\times\left(5-3\right)}{2^{15}\times3^8+2^{13}\times\left(-3\right)^9}\)
\(=\dfrac{3^8\times2^{12}\times2}{2^{13}\times3^8\times\left(4-3\right)}=\dfrac{1}{1}=1\)
\(#PaooNqoccc\)
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a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
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\(\dfrac{5\times4^{15}\times9^9-4\times3^{20}\times8^9}{5\times2^{10}\times6^{19}-7\times2^{29}\times27^6}\\ =\dfrac{5\times2^{30}\times3^{18}-2^2\times3^{20}\times2^{27}}{5\times2^{10}\times3^{19}\times2^{19}-7\times2^{29}\times3^{18}}\\ =\dfrac{5\times2^{30}\times3^{18}-2^{29}\times3^{20}}{5\times2^{29}\times3^{19}-7\times2^{29}\times3^{18}}\\ =\dfrac{2^{29}\times3^{18}\times\left(5\times2-3^2\right)}{2^{29}\times3^{18}\times\left(5\times3-7\right)}\\ =\dfrac{10-9}{15-7}\\ =\dfrac{1}{8}\)
\(A=\dfrac{2^{13}.9^5}{6^8.8^3}=\dfrac{2^{13}.3^{10}}{3^8.2^8.2^9}=\dfrac{2^{13}.3^{10}}{3^8.2^{17}}=\dfrac{3^2}{2^4}=\dfrac{9}{16}\)